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Here are two questions:

(1) Let $a, b$ and $c$ be three fixed positive real numbers, if

$$ \begin{split} \sqrt{a^2-x^2}+\sqrt{c^2-x^2}+\sqrt{a^2-y^2}+\sqrt{b^2-y^2}>\sqrt{b^2-z^2}+\sqrt{c^2-z^2}\\ \sqrt{a^2-x^2}+\sqrt{c^2-x^2}+\sqrt{b^2-z^2}+\sqrt{c^2-z^2}>\sqrt{a^2-y^2}+\sqrt{b^2-y^2}\\ \sqrt{b^2-z^2}+\sqrt{c^2-z^2}+\sqrt{a^2-y^2}+\sqrt{b^2-y^2}>\sqrt{a^2-x^2}+\sqrt{c^2-x^2} \end{split} $$ then find the maximum value of $$ \sqrt{a^2-x^2}+\sqrt{a^2-y^2}+\sqrt{b^2-y^2}+\sqrt{b^2-z^2}+\sqrt{c^2-z^2}+\sqrt{c^2-x^2}\qquad (*) $$ where $x,y,z>0$.

I konw that if $x=y=z$, $(*)$ takes maximum, but I can't prove it without partial derivative. Is there some good ideas to solve it by the skill of inequality?

Mr.He
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1 Answers1

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Let $$\sqrt{a^2-x^2}+\sqrt{c^2-x^2}+\sqrt{a^2-y^2}+\sqrt{b^2-y^2}>\sqrt{b^2-z^2}+\sqrt{c^2-z^2}\tag1$$ $$\sqrt{a^2-x^2}+\sqrt{c^2-x^2}+\sqrt{b^2-z^2}+\sqrt{c^2-z^2}>\sqrt{a^2-y^2}+\sqrt{b^2-y^2}\tag2$$ $$\sqrt{b^2-z^2}+\sqrt{c^2-z^2}+\sqrt{a^2-y^2}+\sqrt{b^2-y^2}>\sqrt{a^2-x^2}+\sqrt{c^2-x^2}\tag3$$

Also, let $$f(x,y,z):=\sqrt{a^2-x^2}+\sqrt{a^2-y^2}+\sqrt{b^2-y^2}+\sqrt{b^2-z^2}+\sqrt{c^2-z^2}+\sqrt{c^2-x^2}$$

We can say the following :

  • For any fixed $(x,y,z)=(X,Y,Z)$ satisfying $(1)(2)(3)$ and $\min (X,Y,Z)=X$, $(x,y,z)=(\frac X2,\frac X2,\frac X2)$ satisfies $(1)(2)(3)$, and $f(X,Y,Z)\lt f(\frac X2,\frac X2,\frac X2)$ holds.

    (The reason is as follows : Since $a\ge Y\ge X\gt\frac X2$, we see that $(x,y,z)=(\frac X2,\frac X2,\frac X2)$ satisfies $(1)$, i.e. $\sqrt{a^2-(\frac X2)^2}\gt 0$. Similarly, it satisfies $(2)(3)$. We have $\sqrt{a^2-X^2}\lt \sqrt{a^2-(\frac X2)^2}, \sqrt{a^2-Y^2}\lt\sqrt{a^2-(\frac X2)^2}$ etc., so $f(X,Y,Z)\lt f(\frac X2,\frac X2,\frac X2)$ holds.)

Similarly, we can say the followings :

  • For any fixed $(x,y,z)=(X,Y,Z)$ satisfying $(1)(2)(3)$ and $\min (X,Y,Z)=Y$, $(x,y,z)=(\frac Y2,\frac Y2,\frac Y2)$ satisfies $(1)(2)(3)$, and $f(X,Y,Z)\lt f(\frac Y2,\frac Y2,\frac Y2)$ holds.

  • For any fixed $(x,y,z)=(X,Y,Z)$ satisfying $(1)(2)(3)$ and $\min (X,Y,Z)=Z$, $(x,y,z)=(\frac Z2,\frac Z2,\frac Z2)$ satisfies $(1)(2)(3)$, and $f(X,Y,Z)\lt f(\frac Z2,\frac Z2,\frac Z2)$ holds.

So, we can say that if the maximum of $f(x,y,z)$ exists, then the maximum of $f(x,y,z)$ is attained when $x=y=z$.

Now, $f(w,w,w)=2\sqrt{a^2-w^2}+2\sqrt{b^2-w^2}+2\sqrt{c^2-w^2}$ is decreasing with $0\lt w\lt\min(a,b,c)$, so we have $$f(w,w,w)\lt f(0,0,0)=2a+2b+2c$$ The maximum of $f(x,y,z)$ does not exist.

mathlove
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